The Gateway to Computer Science Excellence
0 votes
12 views

If $x$ is real, the set of real values of $a$ for which the function $$y=x^2-ax+1-2a^2$$ is always greater than zero is

  1. $- \frac{2}{3} < a \leq \frac{2}{3}$
  2. $- \frac{2}{3} \leq  a < \frac{2}{3}$
  3. $- \frac{2}{3} < a < \frac{2}{3}$
  4. None of these
in Calculus by Veteran (431k points)
recategorized by | 12 views
0
I have solved this by finding Discriminant.

and getting $x = -2a, \;\text {and} \;2a$

Is this the right method?

Please log in or register to answer this question.

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,291 answers
198,209 comments
104,889 users